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3 years ago

Please help! will give the crown to whoever helps

Mathematics

1 answer:

andrew11 [14]3 years ago

5 0

DEF is congruent to D’E’F’ because DEF is the same shape and size as D’E’F’

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Work out the value of 4bc-b when b=-3 and c=2

Pavel [41]

Answer:

-18

Step-by-step explanation:

4bc-b

Let b = -3 and c=2

4 * (-3)* 2 - (-3)

-24 +6

-18

7 0

4 years ago

Read 2 more answers

Evaluate the limits<br><br>

julsineya [31]

$x > \ln(x)$ for all $x$ , so

$\displaystyle \lim_{x\to\infty} (\ln(x) - x) = - \lim_{x\to\infty} x = \boxed{-\infty}$

Similarly, $\displaystyle \lim_{x\to\infty} (x-e^x) = - \lim_{x\to\infty} e^x = \boxed{-\infty}$

We can of course see the limits are identical by replacing $x\mapsto e^x$ , so that

$\displaystyle \lim_{x\to\infty} (\ln(x) - x) = \lim_{x\to\infty} (\ln(e^x) - e^x) = \lim_{x\to\infty} (x - e^x)$

You can also rewrite the limands to accommodate the application of l'Hôpital's rule. For instance,

$\displaystyle \lim_{x\to\infty} (x - e^x) = \ln\left(\exp\left(\lim_{x\to\infty} (x - e^x)\right)\right) = \ln\left(\lim_{x\to\infty} e^{x-e^x}\right) = \ln\left(\lim_{x\to\infty} \frac{e^x}{e^{e^x}}\right)$

Using the rule, the limit here is

$\displaystyle \lim_{x\to\infty} \frac{(e^x)'}{\left(e^{e^x}\right)'} = \lim_{x\to\infty} \frac{e^x}{e^x e^{e^x}} = \lim_{x\to\infty} \frac1{e^{e^x}} = 0$

so the overall limit is

$\displaystyle \lim_{x\to\infty} (x - e^x) = \ln\left(\lim_{x\to\infty} \frac{e^x}{e^{e^x}}\right) = \ln(0) = \lim_{x\to0^+} \ln(x) = -\infty$

6 0

2 years ago

Ch statement is true about the graphs of the two lines y=-6 and x =

jarptica [38.1K]

The slope answer for this question is 46% rad

8 0

2 years ago

Write the equation of the line that is perpendicular to the line y = 2x + 2 and passes through the point (6, 3).

Mama L [17]

For this case we have that by definition, the equation of the line in a slope-intersection form is given by:

$y = mx + b$

Where:

m: It's the slope

b: It is the cut-off point with the y axis

We have the following equation:

$y = 2x + 2$

Thus, the slope is $m_ {1} = 2$

By definition, if two lines are perpendicular then the product of the slopes is -1.

$m_ {1} * m_ {2} = - 1$

We find $m_ {2}:$

$m_ {2} = \frac {-1} {m_ {1}}\\m_ {2} = \frac {-1} {2}$

Thus, the equation of the line is:

$y = - \frac {1} {2} x + b$

We substitute the given point to find "b":

$3 = - \frac {1} {2} 6 + b\\3 = -3 + b\\b = 6$

Thus, the equation of a line perpendicular to the given line and passing through the given point is:

$y = - \frac {1} {2} x + 6$

Answer:

$y = - \frac {1} {2} x + 6$

7 0

3 years ago

Read 2 more answers

Rewrite the expression as an algebraic expression in x.<br> tan(sin−1(x))

solmaris [256]

Let $tan^{-1} x=\alpha$
Then it can be shown by using a right-angled triangle and the rule of Pythagoras that $\alpha= tan^{-1} \frac{x}{ \sqrt{1-x^{2}} }$
Therefore
$tan(sin ^{-1} (x))=\frac{x}{ \sqrt{1-x^{2}} }$

7 0

4 years ago